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3 years ago

How many grams of iron, Fe, are reacted to form 16.0 grams of FeS (87.91 g/mol)

Chemistry

2 answers:

Pavlova-9 [17]3 years ago

8 0

Answer is: 10.164 grams of iron.

Balanced chemical reaction: 8Fe + S₈ → 8FeS.

m(FeS) = 16.0 g; mass of the compound.

M(FeS) = 87.91 g/mol; molar mass of the compound.

n(FeS) = m(FeS) ÷ M(FeS).

n(FeS) = 16 g ÷ 87.91 g/mol.

n(FeS) = 0.182 mol; amount of substance.

From chemical reaction: n(FeS) : n(Fe) = 8 : 8 (1 : 1).

n(Fe) = 0.182 mol.

m(Fe) = n(Fe) · M(Fe).

m(Fe) = 0.182 mol · 55.845 g/mol.

m(Fe) = 10.164 g.

shtirl [24]3 years ago

4 0

the complete question and answer in the attached figure

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2 years ago

If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

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$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

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The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

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2 years ago

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3 years ago

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1 year ago

Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b.

Ilia_Sergeevich [38]

Answer:

a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe

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In this case, we have to remember which peptidic bonds can break each protease:

-) Trypsin

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With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).

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3 years ago